贡献者： 待更新
电磁场的动量为
\begin{equation}
\boldsymbol{\mathbf{p}} = \epsilon_0 \int \,\mathrm{d}{V} \boldsymbol{\mathbf{E}} \boldsymbol\times \boldsymbol{\mathbf{B}} ~.
\end{equation}
角动量为
\begin{equation}
\boldsymbol{\mathbf{J}} = \boldsymbol{\mathbf{r}} \boldsymbol\times \boldsymbol{\mathbf{p}} = \epsilon_0 \int \,\mathrm{d}{V} \boldsymbol{\mathbf{r}} \boldsymbol\times \boldsymbol{\mathbf{E}} \boldsymbol\times ( \boldsymbol{\nabla}\boldsymbol{\times} \boldsymbol{\mathbf{A}} )~.
\end{equation}
现在假设电磁场只在一定范围内不为零，且体积分的边界处场强为零。假设该范围内没有净电荷与电流，则
\begin{equation} \begin{aligned}
\boldsymbol{\mathbf{r}} \boldsymbol\times \boldsymbol{\mathbf{E}} \boldsymbol\times ( \boldsymbol{\nabla}\boldsymbol{\times} \boldsymbol{\mathbf{A}} ) &= \boldsymbol{\mathbf{r}} \boldsymbol\times [ \boldsymbol\nabla ( \boldsymbol{\mathbf{E}} \boldsymbol\cdot \boldsymbol{\mathbf{A}} ) - \boldsymbol{\mathbf{A}} ( \boldsymbol{\mathbf{E}} \boldsymbol\cdot \boldsymbol{\mathbf{\nabla}} )]_{\partial A} \\
&= [( \boldsymbol{\mathbf{r}} \boldsymbol\times \boldsymbol{\mathbf{\nabla}} )( \boldsymbol{\mathbf{E}} \boldsymbol\cdot \boldsymbol{\mathbf{A}} ) - ( \boldsymbol{\mathbf{E}} \boldsymbol\cdot \boldsymbol{\mathbf{\nabla}} )( \boldsymbol{\mathbf{r}} \boldsymbol\times \boldsymbol{\mathbf{A}} )]_{\partial A}~.
\end{aligned} \end{equation}
其中转微分算符 $[\,]_{\partial A}$ 的作用是先把方括号内的 $ \boldsymbol{\mathbf{\nabla}} $ 作为普通矢量进行计算，再把展开结果中每一项的偏微分作用在 $A$ 的分量上。上式第一项为 $\sum\limits_i {E_i} ( \boldsymbol{\mathbf{r}} \boldsymbol\times \boldsymbol{\mathbf{\nabla}} ) A_i$，第二项为
\begin{equation}
-[( \boldsymbol{\mathbf{E}} \boldsymbol\cdot \boldsymbol{\mathbf{\nabla}} )( \boldsymbol{\mathbf{r}} \boldsymbol\times \boldsymbol{\mathbf{A}} )]_{\partial A} = - ( \boldsymbol{\mathbf{E}} \boldsymbol\cdot \boldsymbol{\mathbf{\nabla}} )( \boldsymbol{\mathbf{r}} \boldsymbol\times \boldsymbol{\mathbf{A}} ) + [( \boldsymbol{\mathbf{E}} \boldsymbol\cdot \boldsymbol{\mathbf{\nabla}} )( \boldsymbol{\mathbf{r}} \boldsymbol\times \boldsymbol{\mathbf{A}} )]_{\partial r}~.
\end{equation}
其中第二项为 $[( \boldsymbol{\mathbf{E}} \boldsymbol\cdot \boldsymbol{\mathbf{\nabla}} )( \boldsymbol{\mathbf{r}} \boldsymbol\times \boldsymbol{\mathbf{A}} )]_{\partial r} = [( \boldsymbol{\mathbf{E}} \boldsymbol\cdot \boldsymbol{\mathbf{\nabla}} ) \boldsymbol{\mathbf{r}} ] \boldsymbol\times \boldsymbol{\mathbf{A}} = \boldsymbol{\mathbf{E}} \boldsymbol\times \boldsymbol{\mathbf{A}} $，第一项中
\begin{equation}
\begin{aligned}
( \boldsymbol{\mathbf{E}} \boldsymbol\cdot \boldsymbol{\mathbf{\nabla}} )( \boldsymbol{\mathbf{r}} \boldsymbol\times \boldsymbol{\mathbf{A}} ) &= [( \boldsymbol{\mathbf{E}} \boldsymbol\cdot \boldsymbol{\mathbf{\nabla}} )( \boldsymbol{\mathbf{r}} \boldsymbol\times \boldsymbol{\mathbf{A}} )]_{\partial ErA} - [( \boldsymbol{\mathbf{E}} \boldsymbol\cdot \boldsymbol{\mathbf{\nabla}} )( \boldsymbol{\mathbf{r}} \boldsymbol\times \boldsymbol{\mathbf{A}} )]_{\partial E}\\
& = [( \boldsymbol{\mathbf{E}} \boldsymbol\cdot \boldsymbol{\mathbf{\nabla}} )( \boldsymbol{\mathbf{r}} \boldsymbol\times \boldsymbol{\mathbf{A}} )]_{\partial ErA}~,
\end{aligned}
\end{equation}
这是因为 $[( \boldsymbol{\mathbf{E}} \boldsymbol\cdot \boldsymbol{\mathbf{\nabla}} )( \boldsymbol{\mathbf{r}} \boldsymbol\times \boldsymbol{\mathbf{A}} )]_{\partial E} = ( \boldsymbol{\nabla}\boldsymbol{\cdot} \boldsymbol{\mathbf{E}} )( \boldsymbol{\mathbf{r}} \boldsymbol\times \boldsymbol{\mathbf{A}} ) = 0$。 综上，
\begin{equation}
\boldsymbol{\mathbf{J}} = \epsilon_0 \int \,\mathrm{d}{V} \sum_i E_i ( \boldsymbol{\mathbf{r}} \boldsymbol\times \boldsymbol{\mathbf{\nabla}} ) A_i + \epsilon_0 \int \,\mathrm{d}{V} \boldsymbol{\mathbf{E}} \boldsymbol\times \boldsymbol{\mathbf{A}} + \epsilon_0 \int \,\mathrm{d}{V} [( \boldsymbol{\mathbf{E}} \boldsymbol\cdot \boldsymbol{\mathbf{\nabla}} )( \boldsymbol{\mathbf{r}} \boldsymbol\times \boldsymbol{\mathbf{A}} )]_{\partial ErA}~.
\end{equation}
现在证明最后一项为 0。以 $x$ 分量为例，
\begin{equation}
\begin{aligned}
\hat{\boldsymbol{\mathbf{x}}} \int \,\mathrm{d}{V} [( \boldsymbol{\mathbf{E}} \boldsymbol\cdot \boldsymbol{\mathbf{\nabla}} )( \boldsymbol{\mathbf{r}} \boldsymbol\times \boldsymbol{\mathbf{A}} )]_{\partial ErA} &= \int \,\mathrm{d}{V} \boldsymbol{\nabla}\boldsymbol{\cdot} [ \boldsymbol{\mathbf{E}} ( \hat{\boldsymbol{\mathbf{x}}} \boldsymbol\cdot \boldsymbol{\mathbf{r}} \boldsymbol\times \boldsymbol{\mathbf{A}} ) ] \\
&= \oint \,\mathrm{d}{ \boldsymbol{\mathbf{s}} } \boldsymbol{\mathbf{E}} ( \hat{\boldsymbol{\mathbf{x}}} \boldsymbol\cdot \boldsymbol{\mathbf{r}} \boldsymbol\times \boldsymbol{\mathbf{A}} ) = 0~.
\end{aligned}
\end{equation}
最后一步是因为边界处场强为零。现在我们可以看出角动量由两部分组成
\begin{equation}
\boldsymbol{\mathbf{J}} = \boldsymbol{\mathbf{L}} + \boldsymbol{\mathbf{S}} ~, \qquad
\boldsymbol{\mathbf{L}} = \epsilon_0 \int \,\mathrm{d}{V} \sum_i E_i ( \boldsymbol{\mathbf{r}} \boldsymbol\times \boldsymbol{\mathbf{\nabla}} ) A_i ~,\qquad
\boldsymbol{\mathbf{S}} = \epsilon_0 \int \,\mathrm{d}{V} \boldsymbol{\mathbf{E}} \boldsymbol\times \boldsymbol{\mathbf{A}} ~.
\end{equation}
其中 $ \boldsymbol{\mathbf{L}} $ 是轨道角动量，$ \boldsymbol{\mathbf{S}} $ 是自旋角动量。